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Question Number 192765 by Mingma last updated on 26/May/23

Answered by ajfour last updated on 26/May/23

Commented by ajfour last updated on 26/May/23

$$\tan \theta =2\left(\frac{1}{3}\right)$$$$r(1+\cos \theta )=\frac{1}{9}$$$$r\sin \theta =h-\frac{1}{3}$$$$\Rightarrow r=\frac{1/9}{1+\frac{3}{\sqrt{13}}}=\frac{13-3\sqrt{13}}{36}$$$$h=\frac{1}{3}+\frac{\sqrt{13}}{9(\sqrt{13}+3)}=\frac{25-3\sqrt{13}}{36}$$$$G=\frac{1}{81}+\frac{1}{2}(2r+r\cos \theta )(h-\frac{1}{3})$$$$-(\pi -\theta )\frac{r^2}{2}$$$$$$

Commented by Mingma last updated on 28/May/23

Nice work!

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